note that 
$$e^{a\partial/\partial x}f(x)=f(x+a)$$
is the translation operator, so your exponent of the delta function gives $2\pi \delta(x-in)$, which is indeed consistent with

$$\int_{-\infty}^\infty e^{izy}dy=2\pi\delta(z)$$

for $z=x-in$.